Find the real values of x and y, if (x + iy) (2 – 3i) = 4 + i

(x + iy) (2 – 3i) = 4 + i

Given:

(x + iy) (2 – 3i) = 4 + i

Let us simplify the expression we get,

x(2 – 3i) + iy(2 – 3i) = 4 + i

2x – 3xi + 2yi – 3yi² = 4 + i

2x + (-3x+2y)i – 3y (-1) = 4 + i [since, i² = -1]

2x + (-3x+2y)i + 3y = 4 + i [since, i² = -1]

(2x+3y) + i(-3x+2y) = 4 + i

Equating Real and Imaginary parts on both sides, we get

2x+3y = 4… (i)

And -3x+2y = 1… (ii)

Multiply (i) by 3 and (ii) by 2 and add

On solving we get,

6x – 6x – 9y + 4y = 12 + 2

13y = 14

y = \(\dfrac{ 14}{13}\)

Substitute the value of y in (i) we get,

2x+3y = 4

2x + 3(\( \dfrac{ 14}{13}\)) = 4

2x = 4 – (\( \dfrac{ 42}{13}\))

= \(\dfrac{ (52-42)}{13}\)

2x = \( \dfrac{10}{13}\)

x = \( \dfrac{5}{13}\)

x = \( \dfrac{5}{13}\), y =\( \dfrac{14}{13}\)

The real values of x and y are \( \dfrac{5}{13}, \dfrac{14}{13}\)